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Explain velocity of charge in combined electric and magnetic field in reference to velocity selector. |
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Answer» Solution :1. A CHARGE q moving with velocity `VECV` in presence of both electric and magnetic fields experiences a force given by, `vecF=vecF_(E)+vecF_(B)` = `vecE_(q)+q(vecv+vecB)""...(1)`  2. Electric and magnetic field are perpendicular to each other and also perpendicular to the velocity of the particle as shown in figure. `vecF_(E)=qvecE=qEhatj""...(2)` and `vecF_(B)=q(vecvxxvecB)=q(vhatixxBhatk)` `vecF_(B)=-qvB(hatj)""...(3)" "(becausehatixxhatk=-hatj)` `thereforevecF=vecF_(E)+vecF_(B)=q[E-VB]hatj` 3. Thus, `vecF_(E)andvecF_(B)` both are opposite to each other 4. Suppose, we adjust the value of `vecEandvecB` such that magnitudes of the two forces are equal. Then, total force on the charge is zero and the charge will move in the fields undeflected. 5. Thus, Eq = qvB `thereforev=E/B""...(4)` 6. This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed `vecEandvecB` fields THEREFORE, serve as a velocity SELECTOR. Only particles with speed `E/B` pass undeflected through the region of crossed fields. 7. This method was employed by J.J. Thomson in 1897 to measure the charge to mass ratio `e/m` of an electron. 8. The principle is also employed in mass spectrometer, a device that separates charged particles, usually ions according to their charge to mass ratio. |
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